How to rename spark data frame output file in AWS in spark SCALA

问题

I am saving my spark data frame output as csv file in scala with partitions. This is how i do that in Zeppelin.

val sqlContext = new org.apache.spark.sql.SQLContext(sc)

    import sqlContext.implicits._
    import org.apache.spark.{ SparkConf, SparkContext }
    import java.sql.{Date, Timestamp}
    import org.apache.spark.sql.Row
    import org.apache.spark.sql.types._
    import org.apache.spark.sql.functions.udf

import org.apache.spark.sql.functions.input_file_name
import org.apache.spark.sql.functions.regexp_extract

val get_cus_val = spark.udf.register("get_cus_val", (filePath: String) => filePath.split("\\.")(3))

val rdd = sc.textFile("s3://trfsmallfffile/FinancialLineItem/MAIN")
val header = rdd.filter(_.contains("LineItem.organizationId")).map(line => line.split("\\|\\^\\|")).first()
val schema = StructType(header.map(cols => StructField(cols.replace(".", "_"), StringType)).toSeq)
val data = sqlContext.createDataFrame(rdd.filter(!_.contains("LineItem.organizationId")).map(line => Row.fromSeq(line.split("\\|\\^\\|").toSeq)), schema)

val schemaHeader = StructType(header.map(cols => StructField(cols.replace(".", "."), StringType)).toSeq)
val dataHeader = sqlContext.createDataFrame(rdd.filter(!_.contains("LineItem.organizationId")).map(line => Row.fromSeq(line.split("\\|\\^\\|").toSeq)), schemaHeader)

val df1resultFinal=data.withColumn("DataPartition", get_cus_val(input_file_name))
val rdd1 = sc.textFile("s3://trfsmallfffile/FinancialLineItem/INCR")
val header1 = rdd1.filter(_.contains("LineItem.organizationId")).map(line => line.split("\\|\\^\\|")).first()
val schema1 = StructType(header1.map(cols => StructField(cols.replace(".", "_"), StringType)).toSeq)
val data1 = sqlContext.createDataFrame(rdd1.filter(!_.contains("LineItem.organizationId")).map(line => Row.fromSeq(line.split("\\|\\^\\|").toSeq)), schema1)


import org.apache.spark.sql.expressions._
val windowSpec = Window.partitionBy("LineItem_organizationId", "LineItem_lineItemId").orderBy($"TimeStamp".cast(LongType).desc) 
val latestForEachKey = data1.withColumn("rank", rank().over(windowSpec)).filter($"rank" === 1).drop("rank", "TimeStamp")


val dfMainOutput = df1resultFinal.join(latestForEachKey, Seq("LineItem_organizationId", "LineItem_lineItemId"), "outer")
      .select($"LineItem_organizationId", $"LineItem_lineItemId",
        when($"DataPartition_1".isNotNull, $"DataPartition_1").otherwise($"DataPartition").as("DataPartition"),
        when($"StatementTypeCode_1".isNotNull, $"StatementTypeCode_1").otherwise($"StatementTypeCode").as("StatementTypeCode"),
        when($"FinancialConceptLocalId_1".isNotNull, $"FinancialConceptLocalId_1").otherwise($"FinancialConceptLocalId").as("FinancialConceptLocalId"),
        when($"FinancialConceptGlobalId_1".isNotNull, $"FinancialConceptGlobalId_1").otherwise($"FinancialConceptGlobalId").as("FinancialConceptGlobalId"),
        when($"FinancialConceptCodeGlobalSecondaryId_1".isNotNull, $"FinancialConceptCodeGlobalSecondaryId_1").otherwise($"FinancialConceptCodeGlobalSecondaryId").as("FinancialConceptCodeGlobalSecondaryId"),
        when($"FFAction_1".isNotNull, $"FFAction_1").otherwise($"FFAction|!|").as("FFAction|!|"))
        .filter(!$"FFAction|!|".contains("D|!|"))

val dfMainOutputFinal = dfMainOutput.na.fill("").select($"DataPartition",$"StatementTypeCode",concat_ws("|^|", dfMainOutput.schema.fieldNames.filter(_ != "DataPartition").map(c => col(c)): _*).as("concatenated"))

val headerColumn = dataHeader.columns.toSeq

val header = headerColumn.mkString("", "|^|", "|!|").dropRight(3)

val dfMainOutputFinalWithoutNull = dfMainOutputFinal.withColumn("concatenated", regexp_replace(col("concatenated"), "|^|null", "")).withColumnRenamed("concatenated", header)


dfMainOutputFinalWithoutNull.repartition(1).write.partitionBy("DataPartition","StatementTypeCode")
  .format("csv")
  .option("nullValue", "")
  .option("delimiter", "\t")
  .option("quote", "\u0000")
  .option("header", "true")
  .option("codec", "gzip")
  .save("s3://trfsmallfffile/FinancialLineItem/output")

  val FFRowCount =dfMainOutputFinalWithoutNull.groupBy("DataPartition","StatementTypeCode").count

  FFRowCount.coalesce(1).write.format("com.databricks.spark.xml")
  .option("rootTag", "FFFileType")
  .option("rowTag", "FFPhysicalFile")
  .save("s3://trfsmallfffile/FinancialLineItem/Descr")

Now files are saved in partitioned folder structure which is expected .

Now my requiremen is to rename all the part file and save it in one directory . The name of the file will be as the name of the folder structure .

For example i have one file saved in folder/DataPartition=Japan/PartitionYear=1971/part-00001-87a61115-92c9-4926-a803-b46315e55a08.c000.csv.gz

Now i want my file name to be

Japan.1971.1.txt.gz
Japan.1971.2.txt.gz

I have done this in java map-reduce after my job is completed then i was reading HDFS files system and then moved it into different location as renamed file name .

But how do to the this in AWS S3 files system in spark SCALA .

As far as i have research there is no direct way to rename spark data frame output file name.

But there is implementation that can be done in the job itself using MultipleOutputs as saveAsHadoopFile but how to do that ?.

I am looking for some sample code in scala

It is as like after completing job we need to read the file from s3,reame it and move it to some other location .

回答1:

val tempOutPath = "mediamath.dir"
headerDf.union(outDf)
  .repartition(1)
  .write
  .mode(SaveMode.Overwrite)
  .format("text")
  .option("codec", "gzip")
  .save(tempOutPath)

import org.apache.hadoop.fs._
val sc = spark.sparkContext
val fs = FileSystem.get(sc.hadoopConfiguration)
val file = fs.globStatus(new Path("mediamath.dir/part*.gz"))(0).getPath.getName

fs.rename(new Path("mediamath.dir/" + file), new Path(<aws-s3-path>))

Here is my code snippet please see if this helps you.

回答2:

AFAIK, If you are looking to rename file/object in S3 bucket directly, It's not possible.

You can achieve the rename = copy to target + delete source

First let's extract the filename from source

def prepareNewFilename(oldFilename: String) = {

  val pattern = raw".*/DataPartition=%s/PartitionYear=%s/part-%s.*\.%s"
    .format("([A-Za-z]+)", "([0-9]+)", "([0-9]+)", "([a-z]+)")
    .r

  val pattern(country, year, part, extn) = oldFilename

  "%s.%s.%s.%s.%s".format(country, year, part, "txt", extn)
} 

val oldFilename = "folder/DataPartition=Japan/PartitionYear=1971/part-00001-87a61115-92c9-4926-a803-b46315e55a08.c000.csv.gz"

val newFilename = prepareNewFilename(oldFilename)
//newFilename: String = Japan.1971.00001.txt.gz

Code to rename the file/object in S3 in bucket

import com.amazonaws.AmazonServiceException
import com.amazonaws.services.s3.AmazonS3ClientBuilder

val s3 = AmazonS3ClientBuilder.defaultClient()

try {
  s3.copyObject(sourceBkt, oldFilename, targetBkt, newFilename)
  s3.deleteObject(sourceBkt, oldFilename)
} catch {
  case e: AmazonServiceException =>
    System.err.println(e.getErrorMessage)
    System.exit(1)
}

来源：https://stackoverflow.com/questions/46703623/how-to-rename-spark-data-frame-output-file-in-aws-in-spark-scala

标签

scala

apache-spark

amazon-s3

spark-dataframe

multipleoutputs